Correct Answer: B. \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)

Explanation

\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)

= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)

Correct Answer: D. 3x² - 5x + 3

Correct Answer: B. 3

Explanation

\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

= \(\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)

= \(\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}\)

= \(\frac{6}{2} = 3\)

Correct Answer: B. 1, 4

Explanation

when y = 0, value of x = roots of the equation

x² - 5x + 4 = 0

x = 1, 4 (see graph)

Explanation

Gradient of BC = \(\frac{8 + 2}{1 - 4} = -\frac{10}{3}\)

Gradient of DC = \(\frac{8 - y}{1 - x}\)

BDC is a straight line , so

Gradient of BC = Gradient of DC.

\(\frac{10}{-3} = \frac{8 - y}{1 - x}\)

\(-3(8 - y) = 10(1 - x) \implies 3y - 24 = 10 - 10x\)

\(3y + 10x = 34\)

(b) \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}\)

= \((i + 8j) - (4i - 2j)\)

= \(-3i + 10j\)

\(|\overrightarrow{BC}| = \sqrt{(-3)^{2} + (10)^{2}}\)

= \(\sqrt{109}\)

Unit vector in the direction of BC = \(\frac{(-3i + 10j)}{\sqrt{109}}\)

Correct Answer: D. 400

Correct Answer: D. 10

Explanation

let the number of question = n

n(n + 50) = 600

n² + 50n - 600 - 0 by quadratic formular

x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\)

n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\)

= \(\frac{-50 \pm \sqrt{4900}}{2}\)

= \(\frac{-50 + 70}{2}\)

= \(\frac{20}{2}\)

= 10

Explanation

(a)(i) \(T_{3} = 2 ; ar^{2} = 2 ... (i)\)

\(T_{6} = 54 ; ar^{5} = 54 ... (ii)\)

\(\frac{T_{6}}{T_{3}} = \frac{ar^{5}}{ar^{2}} = \frac{54}{2}\)

\(r^{3} = 27 \implies r = 3 \)

(ii) First term = a

\(T_{3} = ar^{2} = 2\)

\(a = \frac{2}{3^{2}} = \frac{2}{9}\)

(iii) \(S_{10} = \frac{a(r^{n} - 1)}{r - 1}\)

= \(\frac{2(3^{10} - 1)}{9(3 - 1)}\)

= \(\frac{3^{10} - 1}{9}\)

= \(\frac{59049 - 1}{9} \approxeq 6561\) (nearest whole number).

(b) \((1 + 2x)^{n} = 1 + ^{n}C_{1} (1)^{n - 1} (2x) + ^{n}C_{2} (1)^{n - 2} (2x)^{2} + ^{n}C_{3} (1)^{n - 3} (2x)^{3} + ^{n}C_{4} (1)^{n - 4} (2x)^{4}+ ...\)

\(\frac{\text{coefficient of} x^{4}}{\text{coefficient of } x^{3}} = \frac{3}{1}\)

\(\frac{n!}{4! (n - 4)!} \times 2^{4} \times \frac{3! (n - 3)!}{n!} = \frac{3}{1}\)

\(\frac{16 \times 3! (n - 3)(n - 4)!}{4 \times 3! (n - 4)! \times 8} = \frac{3}{1}\)

\(\frac{2(n - 3)}{4} = \frac{3}{1} \implies 2n - 6 = 12\)

\(2n = 18 \implies n = 9\)

Correct Answer: C. Joseph is a Nigerian footballer therefore he is a good footballer

Explanation

Nigeria footballers from a subset of good footballers.

Correct Answer: B. 5/2